Download - What can scala puzzlers teach us
What Can Scala Puzzlers Teach Us?Daniel C. Sobral
Disclaimer #1: "Puzzlers"
● Java Puzzlers (the book/presentations):○ Weird behavior in Java that is difficult to predict or
explain● This presentation:
○ Weird behavior in Scala that confounds beginners■ (and sometimes experienced programmers)
○ FAQs, not Fun
I can't speak for any of the people who contributed to create and turn Scala into what it is. Furthermore, if you learn anything from this presentation, learn that a perfect understanding of Scala is an unlikely achievement...
Disclaimer #2: don't trust me.
Symbols, operators and punctuation
● What does it mean?● Where can I find information about it?● WTF?
The symbol puzzler
Source: Stack Overflow
The Origin of the Symbols
● Punctuation elements of the language○ @ # ( ) [ ] { } , . ; : ` ' "
● Keywords, reserved symbols and XML○ <- => <: >: <% // /* */ _* <? <!
● Normal definitions from libraries○ <:< =:= ☆ ★ η :/
● Underscore○ All of the above!
Symbols may be part of the language or come from external libraries; it can be difficult to tell where they begin and end; they are difficult to search for on the Internet, non-mnemonic and unpronounceable.
Then why?
def f(xs: Array[Int], a: Int, b: Int) =
for (i <- xs.indices)
xs(i) = a * i + b
So that libraries can extend the language seamlessly
def f(xs: M[T], a: T, b: T) =
for (i <- xs.indices)
xs(i) = a * i + bFor M and T library-types that support these operations
Which still doesn't really justify unicode...
Advice regarding Symbols
● Learn the language:○ What symbols are "built-in"○ The syntax rules
● If exploring code using unknown libraries,○ Use an IDE to explore it
harsh, indeed...
implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {
new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {
def id[A] = ☆(_ η)
def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g
}
}
Let's try to figure this one out
implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {
new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {
def id[A] = ☆(_ η)
def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g
}
}
Split Scala Syntax from Identifiers
implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {
new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {
def id[A] = ☆(_ η)
def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g
}
}
Identify what we are defining
implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {
new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {
def id[A] = ☆(_ η)
def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g
}
}
Where we are using our definitions
implicit def KleisliCategory[M[_] : Monad]: Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] = {
new Category[({type λ[α, β] = Kleisli[M, α, β]})#λ] {
def id[A] = ☆(_ η)
def compose[X, Y, Z](f: Kleisli[M, Y, Z], g: Kleisli[M, X, Y]) = f <=< g
}
}
What's left are outside identifiers
The Underscore
● What does it mean?● Why does it work here but not here?● WTF?● It's everywhere! It's the borg operator!
One underscore puzzler
Source: Stack overflow
The meanings of underscore
Source: myself, by way of Stack Overflow
The underscore is Scala's wildcard; it means "something" or "anything" and even "nothing". However, it's semantics can be widely different, even in similar contexts.
Why?
● ???● Picking a different symbol for each case
would stretch ASCII● Picking keywords is not in the spirit of Scala● They are somewhat intuitive
○ except when they bite you
Advice regarding underscore
● Intuition does go a long way● But, again, learn the language● And, in particular, know the difference
between these:○ f _○ f(_, y)○ f(x + _, y)
Eta, partial function application and placeholders
● f _○ an eta expansion: turn method f into an anonymous function.
● f(_, y)○ a partial function application: creates an anonymous function from
method f, whose parameters are the ones wholly replaced by underscores.
● f(x + _, y)○ a placeholder: creates an anonymous function from the expression "x
+ _", whose parameters are the underscores (and that function gets passed to f).
_ always picks the tightest non-degenerate scope it can
Seth Tisue
Erasure
● Where are my types?● How do I get them back?● WTF?
Some things that don't work
trait X {
def overload(f: Int => Int)
def overload(f: String => String)
}
Some things that don't work
def f[T](list: List[T]) = list match {
case _: List[Int] => "Ints"
case _: List[String] => "Strings"
case _ => "???"
}
Some things that don't work
class C[T] {
def f(obj: Any) = obj match {
case _: T => "Ours"
case _ => "Not Ours"
}
}
Why?
● JVM does not support "type parameters"● If Scala added metadata to "eliminate"
erasure, the interaction gap between it and Java would be much greater○ Though things like implicits and default parameters
already do it, to some extent○ At least, implicits and defaults are easy to avoid
Some things that don't work
trait X {
def overload(f: Function1[Int, Int])
def overload(f: Function1[String, String])
}
Erasure at work
trait X {
def overload(f: Function1)
def overload(f: Function1)
}
Erasure at work
def f(list: List) = list match {
case _: List => "Ints"
case _: List => "Strings"
case _ => "???"
}
Erasure at work
class C {
def f(obj: Any) = obj match {
case _: Any => "Ours"
case _ => "Not Ours"
}
}
A partial solution
class C[T : ClassTag] {
def f(obj: Any) = obj match {
case _: T => "Ours"
case _ => "Not Ours"
}
} though it will still ignore T's type parameters
Initialization Ordertrait A {
val a: Int
val b = a * 2
}
class B extends A {
val a = 5
println(b)
}
new B
All val’s are initialized in the order they are found.
Initialization Ordertrait A { val foo: Int val bar = 10 println("In A: foo: " + foo + ", bar: " + bar)}
class B extends A { val foo: Int = 25 println("In B: foo: " + foo + ", bar: " + bar)}
class C extends B { override val bar = 99 println("In C: foo: " + foo + ", bar: " + bar)}
new C
Source: Paul Phillips scala-faq by way of scala puzzlers
Initialization OrderIn A: foo: 0, bar: 0
In B: foo: 25, bar: 0
In C: foo: 25, bar: 99
Initialization Ordertrait A { val foo: Int val bar = 10 println("In A: foo: " + foo + ", bar: " + bar)}
class B extends A { val foo: Int = 25 println("In B: foo: " + foo + ", bar: " + bar)}
class C extends B { override val bar = 99 println("In C: foo: " + foo + ", bar: " + bar)}
new C
A val is initialized only once.
Initialization Ordertrait A { val x = 5 }
trait B {
val x: Int
println(x * 2)
}
trait C extends A with B
trait D extends B
class E extends D with C { println(x * 2) }
class F extends C with D { println(x * 2) }
new E
new F
Initialization Order0
10
10
10
Initialization Ordertrait A { val x = 5 }
trait B {
val x: Int
println(x * 2)
}
trait C extends A with B
trait D extends B
class E extends B with D with A with B with C { println(x * 2) }
class F extends A with B with C with B with D { println(x * 2) }
new E
new F
Trait linearization...
Advice regarding initialization order
● Learn the _________
Other things to look out for
● Pattern matching on val's and for's● All sorts of things implicit● Syntactic sugars, auto-tupling and the like● The return keyword
So, what did I learn?
It seems to me that there's a wide range of Scala features that depend on the user having deep knowledge of the language when something goes wrong. "What's going on?" seems to be much more common than "How do I do this?"
As for real Puzzlers...● Site:
○ http://scalapuzzlers.com/● Book:
○ http://www.artima.com/shop/scala_puzzlers